For \(g\isin\Z\) , let \(\bar{g}\isin\Z_{37}\) denote the residue class of g modulo 37. Consider the group U 37 = { \(\bar{g}\isin\Z_{37}:1\le g\le37\) with gcd(g, 37) = 1} with respect to multiplication modulo 37. Then which one of the following is FALSE?

The set \({\bar g \isin U_{37}: \bar g = (\bar g)^{-1}}\) contains exactly 2 elements.

The order of the element \(\overline{10}\) in U 37 is 36.

There is exactly one group homomorphism from U 37 to ( \(\Z\) , +).

There is exactly one group homomorphism from U 37 to ( \(\mathbb{Q}\) ,+).

The order of the element \(\overline{10}\) in U 37 is 36.

The set \({\bar g \isin U_{37}: \bar g = (\bar g)^{-1}}\) contains exactly 2 elements.

The order of the element \(\overline{10}\) in U 37 is 36.

There is exactly one group homomorphism from U 37 to ( \(\Z\) , +).

There is exactly one group homomorphism from U 37 to ( \(\mathbb{Q}\) ,+).

Question:

For \(g\isin\Z\), let \(\bar{g}\isin\Z_{37}\) denote the residue class of g modulo 37. Consider the group U₃₇ = {\(\bar{g}\isin\Z_{37}:1\le g\le37\) with gcd(g, 37) = 1} with respect to multiplication modulo 37. Then which one of the following is FALSE?

Updated On: Oct 1, 2024

The set \({\bar g \isin U_{37}: \bar g = (\bar g)^{-1}}\) contains exactly 2 elements.
The order of the element \(\overline{10}\) in U₃₇ is 36.
There is exactly one group homomorphism from U₃₇ to (\(\Z\), +).
There is exactly one group homomorphism from U₃₇ to (\(\mathbb{Q}\),+).

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The correct option is (B): The order of the element \(\overline{10}\) in U₃₇ is 36.

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